Ask question

Ask question

All categories

3 years ago

11

What are real images? Are the upright or inverted?

1 answer:

Mars2501 [29]3 years ago

6 0

Answer:

A real image can be viewed when a screen is placed in the plane of convergence. Real images are inverted... HOPE THIS HELPS!!!

Explanation:

You might be interested in

Where is potential energy in food stored

AlexFokin [52]

The potential energy is stored in the chemical bonds of the food. When those bonds break up during the metabolic processes, the energy is released. After that, that energy is stored in the Adenosine Triphosphate bonds aka ATP. The simplest way to think is to think of food as the tightly bound atoms. When the chemical bonds between those atoms break, the stored energy in that food is released.

6 0

3 years ago

A space shuttle travels around the Earth at a constant speed of 28000 kilometers per hour. If it takes 90 minutes to complete on

riadik2000 [5.3K]

Answer:

Explanation:

14

8 0

3 years ago

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

or

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0

3 years ago

We know the frequency range of certain sounds are: 400-560 Hz, what are the ranges of wavelength in meters when the signal trans

Ksivusya [100]

Answer:

Range of wavelength will be $5.35\times 10^5m$ to $7.5\times 10^5m$

Explanation:

We have given range of frequency is 400-560 Hz

Speed of the light $c=3\times 10^8m/sec$

We have to find the range of the wavelength of signal transmitted

Ween know that velocity is given by $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency

So for 400 Hz frequency wavelength will be $\lambda =\frac{c}{f}=\frac{3\times 10^8}{400}=7.5\times 10^5m$

And wavelength for frequency 560 Hz $\lambda =\frac{c}{f}=\frac{3\times 10^8}{560}=5.35\times 10^5m$

So range of wavelength will be $5.35\times 10^5m$ to $7.5\times 10^5m$

8 0

3 years ago

a 0.11kg moving hockey puck is caught by a 80kg goalie who then slides on the frictionless ice at 0.076 m/s. What was the initia

aleksandrvk [35]

Answer:

55mph

Explanation:

3 0

3 years ago